What are the sources of family resemblance? How do we quantify them numerically?
Heritability
Proportion of similarity in phenotypes that can be attributed to similarity in genotypes.
Model: Phenotype (P) = Genotype (G) + Environment (E)Variance decomposition \[\mathrm{var}(P) = \mathrm{var}(𝐺) + \mathrm{var}(𝐸)\] Proportion of variance \[h^2 = \frac{\mathrm{var}(𝐺)}{\mathrm{var}(𝑃)}\]
Tenesa, A., Haley, C. The heritability of human disease: estimation, uses and abuses . Nat Rev Genet 14 , 139–149 (2013). doi:10.1038/nrg3377
Visscher, P., Hill, W. & Wray, N. Heritability in the genomics era — concepts and misconceptions . Nat Rev Genet 9 , 255–266 (2008). doi:10.1038/nrg2322
How to estimate heritability from data
Plot of child (offspring) height versus the average of their parents’ heights. What is a statistic that can be used to summarise the relationship between these two variables?
How to estimate heritability from data
\(\beta = \frac{\mathrm{cov}(A, B)}{\mathrm{var}(A)}\)
Estimate the beta coefficient (slope) for a simple regression from the covariance between predictor (\(A\) ) and outcome (\(B\) ) variable divided by the variance of the predictor (\(A\) ).
Simple model of genetic and environmental effects
\[
P = G + E
\]
The phenotype value \(P\) is influenced by a genetic effect \(G\) and and environmental effect \(E\) .
For simplicity assume that \(P\) is an individual’s deviation from the average phenotype in the population.
Simple model of genomics
\[
G = d + s
\]
Each individual has two copies of the genome, one inherited from each parent.
Here we label the genome from the mother \(d\) and the genome from the father \(s\) . These are the transmitted genetic effects.
Simple model of inheritance
Each parent also has a copy of the genome which they do not pass on to their child, labelled \(d^\prime\) for the mother and \(s^\prime\) for the father. These are the non-transmitted genetic effects.
Simple model of genetics, environment, and inheritance
Phenotype (\(P\) ) value is the sum of the two genetic values plus an environmental value (\(e\) ).
Mother’s phenotype: \(P_d = d + d^\prime + e_d\)
Father’s phenotype: \(P_s = s + s^\prime + e_s\)
Child’s phenotype: \(P_o = d + s + e_o\)
Regression equation
\(\beta = \frac{\mathrm{cov}(A, B)}{\mathrm{var}(A)}\)
\(A\) = average of parents’ phenotypes\(B\) = offspring phenotype
Therefore, \(\beta = \frac{\mathrm{cov}(\frac{P_d + P_s}{2}, P_o)}{\mathrm{var}(\frac{P_d + P_s}{2})}\)
Parent–offspring covariance
\[
\mathrm{cov}(\frac{P_d + P_s}{2}, P_o)
\]
\[
= \mathrm{cov}(\frac{d + d^\prime + e_d + s + s^\prime + e_s}{2}, d + s + e_o)
\]
Parent-offspring covariance
Expand the terms. Recall that:
\[
\mathrm{cov}(A+X,B+Y) = \\
\mathrm{cov}(A,B) + \mathrm{cov}(A,Y) + \mathrm{cov}(X,B) + \mathrm{cov}(X,Y)
\] Thus we can do a pairwise expansion to: \[
= \mathrm{cov}(\frac{d}{2} + \frac{d^\prime}{2} + \frac{e_d}{2} + \frac{s}{2} + \frac{s^\prime}{2} + \frac{e_s}{2}, d + s + e_o)
\] \[
= \mathrm{cov}(\frac{d}{2}, d) + \mathrm{cov}(\frac{d^\prime}{2}, d) + \dotsm+ \mathrm{cov}(\frac{e_s}{2}, e_o)\] $$
Simplifications
Some terms can be simplified.
Covariance between a genetic effect and itself \[
\mathrm{cov}(\frac{d}{2}, d), \mathrm{cov}(\frac{s}{2}, s)
\]
Simplifies to:
\[
\mathrm{cov}(\frac{d}{2}, d) = \frac{1}{2}\mathrm{cov}(d, d) = \frac{1}{2}\mathrm{var}(d)
\] \[
\mathrm{cov}(\frac{s}{2}, s) = \frac{1}{2}\mathrm{cov}(s, s) = \frac{1}{2}\mathrm{var}(s)
\]
Assumptions
For some terms we might make an assumption that they are equal to 0.
Covariance between genetic effects from the same parent \[
\mathrm{cov}(\frac{d^\prime}{2}, d), \mathrm{cov}(\frac{s^\prime}{2}, s)
\]
If there is inbreeding (individual does not have ancestors that are closely related) or assortative mating in the grandparental generation, then genetic effects from the same parent would be expected to have a non-zero covariance.
Covariance between genetic effects from different parents \[
\mathrm{cov}(\frac{d^\prime}{2}, s), \mathrm{cov}(\frac{s^\prime}{2}, d)
\]
If there is assortative mating between the parents, then these effects would be expected to covary
Covariance between parent and offspring environment effects \[
\mathrm{cov}(\frac{e_d}{2}, e_o), \mathrm{cov}(\frac{e_s}{2}, e_o)
\]
If aspects of the parental environment are transmitted as well or if parents and offspring tend to encounter similar environments, then these effects are expected to covary.
Covariance between parental genetic and offspring environmental effects \[
\mathrm{cov}(\frac{d}{2}, e_o), \mathrm{cov}(\frac{s}{2}, e_o)
\]
If parental genetic effects shape the environment of the offspring (referred to as indirect genetic effects), then these effects are expected to covary as well.
Using those assumptions the parent–offspring covariance simplifies to
\[
\mathrm{cov}(\frac{P_d + P_s}{2}, P_o) = \frac{\mathrm{var}(d) + \mathrm{var}(s)}{2}
\]
Parent variance
The denominator in the regression equation was \[
\mathrm{var}(\frac{P_d + P_s}{2})
\]
Using the identity \[
\mathrm{var}(aX + bY) = a^2\mathrm{var}(X) + b^2\mathrm{var}(Y) + 2ab\mathrm{cov}(X, Y)
\] the variance of the average parental phenotypes is: \[
\mathrm{var}(\frac{P_d + P_s}{2}) = \mathrm{var}(\frac{1}{2}P_d + \frac{1}{2} P_s)
\] \[
= \left(\frac{1}{2}\right)^2\mathrm{var}(P_d) + \left(\frac{1}{2}\right)^2\mathrm{var}(P_s) + 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \mathrm{cov}(P_d, P_s)
\] \[
= \frac{1}{4}\mathrm{var}(P_d) + \frac{1}{4}\mathrm{var}(P_s) + \frac{1}{2} \mathrm{cov}(P_d, P_s)
\]
If we assume as above that there is no covariation between parental effects (\(\mathrm{cov}(P_d, P_s) = 0\) ), this simplifies to
\[
= \frac{\mathrm{var}(P_d) + \mathrm{var}(P_s)}{4}
\]
Thus the regression equation is:
\[
\beta = \frac{\mathrm{cov}(\frac{P_d + P_s}{2}, P_o)}{\mathrm{var}(\frac{P_d + P_s}{2})} \\
= \frac{\frac{\mathrm{var}(d) + \mathrm{var}(s)}{2}}{\frac{\mathrm{var}(P_d) + \mathrm{var}(P_s)}{4}} \\
= 2\frac{\mathrm{var}(d) + \mathrm{var}(s)}{\mathrm{var}(P_d) + \mathrm{var}(P_s)}
\]
Previously we defined
\[
G = d + s
\] thus \[
\mathrm{var}(G) = \mathrm{var}(d) + \mathrm{var}(s)
\] and assume variances in parental phenotypes are equal \[
\mathrm{var}(P_d) = \mathrm{var}(P_s) = \mathrm{var}(P)
\]
Then substitute into the regression equation
\[
\beta = 2\frac{\mathrm{var}(d) + \mathrm{var}(s)}{\mathrm{var}(P_d) + \mathrm{var}(P_s)} \\
= 2 \frac{\mathrm{var}(G)}{\mathrm{var}(P) + \mathrm{var}(P)} \\
= 2 \frac{\mathrm{var}(G)}{2 \mathrm{var}(P)} \\
= \frac{\mathrm{var}(G)}{\mathrm{var}(P)} \\
= h^2
\]
In other words, the beta coefficient from the regression of offspring phenotype on average parent phenotype yields an estimate of the heritability!
Height data
Using the height data, we can calculate the mid-parent–offspring covariance and mid-parent variance and then use that as an estimate for the heritability.
\(\mathrm{cov}(\frac{P_d + P_s}{2}, P_o) =\) 12.57\(\mathrm{var}(\frac{P_d + P_s}{2}) =\) 22.04\(\hat{h}^2 =\) 12.57 / 22.04 = 0.57
Parent and offspring phenotypes become more highly correlated as heritability increases.
Mini review: What assumptions have we made when estimating \(h^2\) ?
Parents’ environments are not similar: \(\mathrm{cov}(e_d, e_s) = 0\)
No assortative mating: \(\mathrm{cov}(d, s) = 0\)
Parents do not transmit their environments: \(\mathrm{cov}(e_o, e_d) = 0\) , \(\mathrm{cov}(e_o, e_s) = 0\)
No gene-environment correlation: \(\mathrm{cov}(d, e_d) = 0\) , \(\mathrm{cov}(s, e_s) = 0\)
No genetic nature: \(\mathrm{cov}(d, e_o) = 0\) , \(\mathrm{cov}(s, e_o) = 0\)
No inbreeding: \(\mathrm{cov}(d, d^\prime) = 0\) , \(\mathrm{cov}(s, s^\prime) = 0\)
Genetic effects are additive: \(Y = a + a^\prime + e\)
Genetic influence is the same for both sexes: \(\mathrm{var}(d + d^\prime) = var(s + s^\prime)\)
Generalising to other relatives
Heritability can also be estimated from resemblance between different types of related pairs. The general equation is:
\[
h^2 = \frac{b}{r}
\]
\(b\) = regression coefficient\(r\) = coefficient of additive variance (“relatedness”)
Example data: depression scores
Correlation of depression scores for different pairs of relatives
Fernandez-Pujals AM et al. (2015) Epidemiology and Heritability of Major Depressive Disorder, Stratified by Age of Onset, Sex, and Illness Course in Generation Scotland: Scottish Family Health Study (GS:SFHS). PLOS ONE 10(11): e0142197. doi:10.1371/journal.pone.0142197
Recurrance risk to relatives
\[
\lambda_\mathrm{R} = \frac{P(\mathrm{affected} | \mathrm{relative affected})}{P(\mathrm{affected in population})} = \frac{K_\mathrm{R}}{K}
\]
Represents how much more likely to are to be affected by a disorder given that a relative is affected, compared to someone from the general population.
Example:
\(K_\mathrm{sib} = P(\mathrm{affected} | \mathrm{sibling affected}) = 0.09\) \(K = P(\mathrm{affected in population}) = 0.02\) \(\frac{K_\mathrm{sib}}{K} = \frac{0.09}{0.02} = 4.5\)
Recurrance risk for schizophrenia
Recurrance risk to relatives for schizophrenia in Sweden, which has a baseline risk of \(K = 0.047\) %.
Lichtenstein, P. et al. Recurrence risks for schizophrenia in a Swedish National Cohort. Psychol. Med. 36, 1417–1425 (2006). doi:10.1017/s0033291706008385
Recurrance risk and heritability
Code \(\mathrm{unaffected} = 0, \mathrm{unaffected} = 1\)
If population prevalence is \(K\) , then phenotypic variance is \(K(1-K)\) (Bernoulli distribution)
\(Y\) = score of individual (proband)\(Y_\mathrm{R}\) = score of relative of probandExpectation: \(E[Y] = E[Y_\mathrm{R}] = K\)
\(K_\mathrm{R} = E[Y_\mathrm{R} | Y = 1]\) Probability that both \(Y\) and \(Y_\mathrm{R}\) = 1: \(E[YY_\mathrm{R}] = K \times K_\mathrm{R}\)
\[
\mathrm{cov}(Y, Y_\mathrm{R}) = E[YY_\mathrm{R}] - E[Y] E[Y_\mathrm{R}] \\
= K \times K_\mathrm{R} - K^2 \\
\]
James, J. W. Frequency in relatives for an all-or-none trait. Ann. Hum. Genet. 35, 47–49 (1971). doi:10.1111/j.1469-1809.1956.tb01377.x
Risch N. Linkage strategies for genetically complex traits. I. Multilocus models. Am J Hum Genet. 1990 Feb;46(2):222-8. PMID: 2301392
\[
\mathrm{cov}(Y, Y_\mathrm{R}) = E[YY_\mathrm{R}] - E[Y] E[Y_\mathrm{R}]
\]
\[
= K \times K_\mathrm{R} - K^2 \\
= K(K_\mathrm{R} - K) \\
= K^2 (\frac{K_\mathrm{R}}{K} - 1) \\
= K^2 (\lambda_\mathrm{R} - 1)
\]
Heritability estimate
\[
h^2 = \frac{\mathrm{cov}_\mathrm{R}}{rV_\mathrm{P}} \\
= \frac{K^2 (\lambda_\mathrm{R} - 1)}{rK(1-K)} \\
= \frac{K (\lambda_\mathrm{R} - 1)}{r(1-K)} \\
\]
Estimating environmental effects
Contrast pairs of relatives that have comparable environmental similarity but different genetic similarity.
Monozygotic (MZ) twins \(r = 1.0\)
Dizygotic (DZ) twins \(r = 0.5\)
Twin correlations
MZ twins: \(t(\mathrm{MZ}) = h^2 + c^2\)
DZ twins: \(t(\mathrm{DZ}) = \frac{1}{2}h^2 + c^2\)
Solve for genetic similarity (\(h^2\) )
\[t(\mathrm{MZ}) = h^2 + [t(\mathrm{DZ}) - \frac{1}{2}h^2]\] \[\frac{1}{2}h^2 = t(\mathrm{MZ}) - t(\mathrm{DZ})\] \[h^2 = 2[t(\mathrm{MZ}) - t(\mathrm{DZ})]\]
Solve for shared environment similarity (\(c^2\) )
\[t(\mathrm{MZ}) = 2[t(\mathrm{MZ}) - t(\mathrm{DZ})] + c^2\] \[t(\mathrm{MZ}) = 2[t(\mathrm{MZ}) - t(\mathrm{DZ})] = c^2\] \[c^2 = t(\mathrm{MZ}) - 2t(\mathrm{MZ}) + 2t(\mathrm{DZ})\] \[c^2 = 2t(\mathrm{DZ}) - t(\mathrm{MZ})\]
